如何使用pandas基于另一列查找数量前3个值

Question

• A, B, C, D, E are games. There are 80 different games
• This dataframe has 5,000 rows

How can I list top 3 games, in terms of amount?

Answer 1

首先将DataFrame.explode用于列表到标量，然后通过DataFrame.drop_duplicates删除重复DataFrame.drop_duplicates ，最后通过Series.value_counts和Series.head获得 top3，因为value_counts默认排序：

top3 = df.explode('Games').drop_duplicates(['Games','Room'])['Games'].value_counts().head(3)

Answer 2

Explode your Games columns (if Games contains real Python list) then drop duplicates (according your side notes) and use value_counts with different parameters according to what you want:

1. Top 3 for all rooms:

>>> df.explode('Games') \
      .drop_duplicates(['Games', 'Rooms']) \
      .value_counts('Games').head(3)
Games
A    2
B    2
C    2
dtype: int64

2. Top 3 per room:

>>> df.explode('Games') \
      .drop_duplicates(['Games', 'Rooms']) \
      .value_counts(['Games', 'Rooms']).head(3)
Games  Rooms
A      North    1
       West     1
B      East     1
dtype: int64

Setup:

data = {'Games': [['A', 'B', 'C'], ['B', 'D'], ['B', 'E'], ['A', 'C'], ['D']],
        'Rooms': ['West', 'East', 'East', 'North', 'South']}
df = pd.DataFrame(data)
print(df)

# Output:
       Games  Rooms
0  [A, B, C]   West
1     [B, D]   East
2     [B, E]   East
3     [A, C]  North
4        [D]  South